Example 38 Two bodies are thrown simultaneously from
the same point. One thrown straight op and the other at an
angle a with the horizontal. Both the bodies have velocity
equal to u. Find the separation between the bodies time.
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Answer:
The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2, then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration:
r
12
=
r
0
(12)
+
v
0
(12)
t+
2
1
w
12
t
2
So,
r
12
=
v
0
(12)
t, (because
w
12
=0 and
r
0
(12)
=0)
or ∣
r
12
∣=∣
v
0
(12)
∣t (1)
But ∣
v
01
∣=∣
v
02
∣=v
0
So, from properties of triangle
v
0(12)
=
v
0
2
+v
0
2
−2v
0
v
0
cos(
2
π
−θ
0
)
Hence, the sought distance ∣
r
12
∣=v
0
2(1−sinθ)
t=22m
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